# Question Video: Instantaneous Rates of Change

Calculus 1: Rate of Change (7 of 10) Change in Volume (using Rate of Change)
Calculus 1: Rate of Change (7 of 10) Change in Volume (using Rate of Change)

### Video Transcript

Find the instantaneous rate of
change of 𝑓 of 𝑥 is equal to the square root of 𝑥 at 𝑥 equals 𝑥 one, which
is greater than zero.

Remember, the instantaneous
rate of change of a function 𝑓 of 𝑥 at a point 𝑥 equals 𝑎 is found by taking
the limit as ℎ approaches zero of the average rate of change function. That’s the limit as ℎ
approaches zero of 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 all over ℎ. In this case, we know that 𝑓
of 𝑥 is equal to the square root of 𝑥, and we want to find the instantaneous
rate of change at 𝑥 equals 𝑥 one. So, we’ll let 𝑎 be equal to 𝑥
one. Let’s substitute what we know
into our formula. We want to compute the limit as
ℎ approaches zero of 𝑓 of 𝑥 one plus ℎ minus 𝑓 of 𝑥 one all over ℎ. We need to find the limit as ℎ
approaches zero of the square root of 𝑥 one plus ℎ minus the square root of 𝑥
one all over ℎ.

Now, we can’t do this with
direct substitution. If we do, we end up dividing by
zero and we know that to be undefined. And so instead, we multiply the
numerator and denominator of the function by the conjugate of the numerator, by
the square root of 𝑥 plus one plus ℎ plus the square root of 𝑥 one. On the denominator, we simply
have ℎ times the square root of 𝑥 one plus ℎ plus the square root of 𝑥
one. Then on the numerator, we have
the square root of 𝑥 one plus ℎ times the square root of 𝑥 one plus ℎ, which
is simply 𝑥 one plus ℎ. Then, we multiply the square
root of 𝑥 one plus ℎ by the square root of 𝑥, and negative the square root of
𝑥 one times the square root of 𝑥 one plus ℎ. When we find their sum, we get
zero.

So, all that’s left to do is to
multiply negative the square root of 𝑥 one by the square root of 𝑥 one. And we simply get negative 𝑥
one. 𝑥 one minus 𝑥 one is
zero. And then, we divide through by
ℎ. And so, this becomes the limit
as ℎ approaches zero of one over the square root of 𝑥 one plus ℎ plus the
square root of 𝑥 one. And we can now evaluate this as
ℎ approaches zero. We’re left with one over the
square root of 𝑥 one plus the square root of 𝑥 one, which is one over two
times the square root of 𝑥 one. The instantaneous rate of
change function of 𝑓 of 𝑥 is equal to the square root of 𝑥 is therefore one
over two times the square root of 𝑥 one.

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